Largest area of a trapezoid in a semicircle

calculusderivativesgeometrymaxima-minima

the question reads:

find the maximum area of a trapezium inscribed in a semicircle(radius= R), if the lower base in on the diameter.

The last statement(in bold) is what confuses me. Does this mean the base HAS to be the diameter? because thats how the solution in the book was given. I was hoping for a more generalized and perhaps intuitive approach to solve this problem. We were supposed to use calculus by the way.

Best Answer

The question only states the lower base has to lie on the diameter. However, you can immediately infer that the top has to be a chord parallel to the diameter. Which means that for any given base less than the diameter, you can just expand the base till it equals the diameter, and the area will be higher than what you started with. This implies that any maximal solution will require the base to be the entire diameter.

You need a convenient parameter to work with. I suggest drawing a radial line from the center of the circle to the top right corner of the trapezoid. If you let the angle between that radial line and the diameter be $\theta$ (measured counterclockwise), then you can easily show the area $A = r^2(1+\cos\theta)\sin\theta$, which you should be able to maximise.